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    <script>
        // 利用递归求100的阶乘

        // 100 * 99 * 98 * 97 * … * 1 = ?
        function jiecheng(n) {
            if (n == 1) {
                return 1
            }
            return n * jiecheng(n - 1)
        }
        console.log(jiecheng(100));
    </script>
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